Question: Simplify; express your answer in exponential form. Assume $q\neq 0, k\neq 0$. $\dfrac{{(q^{-4})^{-4}}}{{(q^{-2}k^{-1})^{-4}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{-4}}$ to the exponent ${-4}$ . Now ${-4 \times -4 = 16}$ , so ${(q^{-4})^{-4} = q^{16}}$ In the denominator, we can use the distributive property of exponents. ${(q^{-2}k^{-1})^{-4} = (q^{-2})^{-4}(k^{-1})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{-4})^{-4}}}{{(q^{-2}k^{-1})^{-4}}} = \dfrac{{q^{16}}}{{q^{8}k^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{16}}}{{q^{8}k^{4}}} = \dfrac{{q^{16}}}{{q^{8}}} \cdot \dfrac{{1}}{{k^{4}}} = q^{{16} - {8}} \cdot k^{- {4}} = q^{8}k^{-4}$.